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Stuck badly on chemistry gas hw question?
2 NACl (s) + 2H2SO4 (l) + MnO2 (s) --> Na2SO4 (s) + MnSO4 (s) + 2H2O (g) + Cl2 (g)
Assume that the gas produced is saturated with water vapor at a partial pessure of 28.7 Hg and that it has a volume of .597 L at 27 C and 755 mm Hg pressure
1) What is the mole fraction of Cl2 in the gas?
2) How many grams of NaCl were used in the experiment, assuming complete reaction?
Im confused about the 28.7 Hg, the hg throws me off i wonder if its supposed to be in mm Hg instead of Hg
1 Answer
- David HLv 41 decade agoFavorite Answer
Yes, 28.7 Hg should be 28.7 mmHg.
The mole fraction of a component of an ideal gas is equal to its partial pressure divided by the total pressure. The mole fraction of water is:
28.7/755 = 0.0380.
The mole fraction of Cl2 would be 1 - 0.0380 = 0.962.
The volume of Cl2 would be 0.962 x 0.597 L = 0.574 L.
You can use the ideal gas law, PV=nRT, to calculate the number of moles of Cl2. The number of moles of NaCl is double the moles of Cl2, so you can get the moles of NaCl. It is then a simple matter of multiplying by the molecular weight of NaCl.
I did not work the entire problem for you, but I hope I got you un-stuck.